Note on Mean Vector Testing for High-Dimensional Dependent Observations

TL;DR

This paper revises the proofs of asymptotic normality for mean vector tests in high-dimensional dependent data, correcting previous errors and clarifying the theoretical foundations of the method.

Contribution

It provides corrected proofs for the asymptotic normality of mean vector tests under dependence, ensuring the validity of the original statistical approach.

Findings

Corrected proofs of asymptotic normality for the test statistics.

Identification and clarification of minor calculation discrepancies.

Reinforcement of the validity of high-dimensional mean testing methods.

Abstract

For the mean vector test in high dimension, Ayyala et al.(2017,153:136-155) proposed new test statistics when the observational vectors are M dependent. Under certain conditions, the test statistics for one-same and two-sample cases were shown to be asymptotically normal. While the test statistics and the asymptotic results are valid, some parts of the proof of asymptotic normality need to be corrected. In this work, we provide corrections to the proofs of their main theorems. We also note a few minor discrepancies in calculations in the publication.

Equations

$${\mathcal{H}}_0:\mu =0\text{mbox}versus{\mathcal{H}}_A:\mu \ne 0.$$

$${\mathcal{H}}_0:{\mu }^{(1)}={\mu }^{(2)}\text{mbox}versus{\mathcal{H}}_A:{\mu }^{(1)}\ne {\mu }^{(2)}.$$

$$p=O(n),M=O(n^{1/8}),\mathrm{tr}(\Gamma (a)\Gamma (b)\Gamma (c)\Gamma (d))=o\{{\mathrm{tr}}^2({\Omega }_n^2)\}\forall a,b,c,d\in \mathcal{M}.$$

$${\mathcal{T}}_n=\frac{{\mathcal{M}}_n}{\sqrt{\widehat{\mathrm{var}\left({\mathcal{M}}_n\right)}}},$$

$${\beta }_n(\mu )\simeq \Phi \left(-z_{\alpha }+\frac{n{\mu }^{\top }\mu }{\sqrt{2\mathrm{tr}({\Omega }_n^2)}}\right)$$

$$\mathrm{Cov}(X_t,X_{t+h})=\Gamma (h).$$

$$\hat{\Gamma }(h)=\frac{1}{n}\sum _{t=1}^{n-h}\left(X_t-{\bar{X}}_n\right){\left(X_{t+h}-{\bar{X}}_n\right)}^{\mathrm{T}},$$

$$\gamma =\left(\begin{array}{c}\mathrm{tr}(\Gamma (0))\\ ⋮\\ \mathrm{tr}(\Gamma (M))\end{array}\right)\in {\mathbb{R}}^{M+1},\text{mbox}and{\hat{\gamma }}_n=\left(\begin{array}{c}\mathrm{tr}(\hat{\Gamma }(0))\\ ⋮\\ \mathrm{tr}(\hat{\Gamma }(M))\end{array}\right)\in {\mathbb{R}}^{M+1}.$$

$$\widehat{\mathrm{tr}\left({\Omega }_n\right)}={\beta }_n^{\mathrm{T}}{\hat{\gamma }}_n$$

$$E({\mathcal{M}}_n)=E\left({\bar{X}}_n^{\mathrm{T}}{\bar{X}}_n-\frac{1}{n}\widehat{\mathrm{tr}\left({\Omega }_n\right)}\right)={\mu }^{\mathrm{T}}\mu .$$

$$\mathrm{var}({\mathcal{M}}_n)=\frac{2}{n^2}\mathrm{tr}\left({\Omega }_n^2\right)+o\left(\frac{1}{n^2}\mathrm{tr}\left({\Omega }_n^2\right)\right).$$

$$T_n=\frac{{\mathcal{M}}_n}{\sqrt{\mathrm{var}\left({\mathcal{M}}_n\right)}}\stackrel{d}{\to }N(0,1).$$

$$T_n=\frac{M_n}{\sqrt{\mathrm{var}\left({\mathcal{M}}_n\right)}}=\frac{{\bar{X}}_n^{\mathrm{T}}{\bar{X}}_n-\frac{1}{n}\mathrm{tr}\left({\Omega }_n\right)}{\sqrt{\mathrm{var}\left({\mathcal{M}}_n\right)}}-\frac{\widehat{\mathrm{tr}\left({\Omega }_n\right)}-\mathrm{tr}\left({\Omega }_n\right)}{n\sqrt{\mathrm{var}\left({\mathcal{M}}_n\right)}}={\Delta }_1-{\Delta }_2$$

$${\Delta }_1=\frac{{\bar{X}}_n^{\mathrm{T}}{\bar{X}}_n-\frac{1}{n}\mathrm{tr}\left({\Omega }_n\right)}{\sqrt{\mathrm{var}\left({\mathcal{M}}_n\right)}}$$

$${\Delta }_2=\frac{\widehat{\mathrm{tr}\left({\Omega }_n\right)}-\mathrm{tr}\left({\Omega }_n\right)}{n\sqrt{\mathrm{var}\left({\mathcal{M}}_n\right)}},$$

$${\mathcal{A}}_{ij}=\frac{1}{n^2}\left[X_i^{\mathrm{T}}{X}_j-\mathrm{tr}\left(\Gamma (i-j)\right)\right],$$

$$B_{ij}$$

$$D_{ij}$$

$$F$$

$${\Delta }_1$$

$${\Delta }_{11}=\frac{{\sum }_{i=1}^{k_n}{\sum }_{j=1}^{k_n}{B}_{ij}}{\sqrt{\mathrm{Var}\left(M_n\right)}}\stackrel{d}{\to }N(0,1)$$

$$Y_i=\frac{1}{w_n-M}\sum _{k=(i-1)w_n+1}^{i{w}_n-M}{X}_k.$$

$$B_{ij}=\sum _{k=(i-1)w_n+1}^{i{w}_n-M}\sum _{l=(j-1)w_n+1}^{j{w}_n-M}{X}_k^{\mathrm{T}}{X}_l=(w_n-M{)}^2{Y}_i^{\mathrm{T}}{Y}_j.$$

$${\varphi }_{nij}$$

$$V_{nj}$$

$$S_{nm}$$

$$\sum _{i\ne j}{B}_{ij}=2S_{n{k}_n}.$$

$$\mathrm{E}\left(S_{n(m+1)}|{\mathcal{F}}_{nm}\right)$$

$$\frac{1}{{\sigma }_n^2}\sum _{j=2}^{k_n}\mathrm{E}\left[V_{nj}^2|{\mathcal{F}}_{n(j-1)}\right]\stackrel{p}{\to }\frac{1}{4}.$$

$$\mathrm{E}\left[V_{nj}^2|{\mathcal{F}}_{n(j-1)}\right]$$

$$\mathrm{E}\left({\eta }_n\right)$$

$$\mathrm{E}\left({\eta }_n^2\right)$$

Full text

1 Introduction

Mean vector testing in high dimension is gaining great attention in the recent past with increasing availability of data sets where the number of variables is greater than the sample size. The traditional Hoteling’s $T^{2}$ test and the more recently developed test statistics assume that the observations are independently and identically distributed. Comparison of mean vector for distributions where the samples are dependent is a relatively understudied problem. The one-sample mean vector test under dependence is defined as follows. The observations $X_{1},\ldots,X_{n}$ are assumed to be $p$-dimensional random vectors with mean $\mu$ and a covariance matrix $\Sigma$ that may not be diagonal. The problem of interest is then to test

[TABLE]

In the two-sample case, $X_{1}^{(1)},\ldots,X_{n_{1}}^{(1)}$ and $X_{1}^{(2)},\ldots,X_{n_{2}}^{(2)}$ are two independent groups of $p$-dimensional observations with mean vectors $\mu^{(1)}$, $\mu^{(2)}$ and covariance matrices $\Sigma^{(1)},\Sigma^{(2)}$, respectively. The hypothesis of interest is that the two population means are equal, viz.

[TABLE]

In (Ayyala et al., 2017), we developed a hypothesis test for (1) when the samples follow an $M$-dependent stationary Gaussian process with mean $\mu$ and autocovariance structure given by $\Gamma(0),\ldots,\Gamma(M)$. The rate of increase with respect to $n$ was assumed to be linear for $p$ ($p=O(n)$) and sublinear for $M$ ($M=O(n^{1/8})$). These assumptions ensure that the number of variables is not increasing faster than the sample size and that there are sufficiently many observations for estimating the autocovariance at all lags. The dependence structure is made sparse by assuming ${\rm tr}(\Gamma(a)\Gamma(b)\Gamma(c)\Gamma(d))=o\{{\rm tr}^{2}(\Omega_{n}^{2})\}$ for any $a,b,c,d\in\mathcal{M}$ where $\mathcal{M}=\{-M,\ldots,M\}$ and the matrix $\Omega_{n}$ is the covariance of $\bar{X}_{n}$ multiplied by the sample size, i.e., $\Omega_{n}=\mathop{\sum}_{h\in\mathcal{M}}n^{-1}(n-|h|)\Gamma(h)$. Summarizing the model, the assumptions are

[TABLE]

The test statistic is based on the Euclidean norm of the sample average, $\bar{X}_{n}^{\top}\bar{X}_{n}$. Define $\mathcal{M}_{n}=\bar{X}_{n}^{\top}\bar{X}_{n}-n^{-1}\widehat{{\rm tr}(\Omega_{n})}$ where $\widehat{{\rm tr}(\Omega_{n})}$ is an unbiased estimator for ${\rm tr}(\Omega_{n})$. Under $H_{0}$, this quantity has expected value $\mu^{\top}\mu$. Using this property, the test statistic is constructed as

[TABLE]

where $\widehat{{\rm var}\left(\mathcal{M}_{n}\right)}$ is a ratio-consistent estimator for ${\rm var}\left(\mathcal{M}_{n}\right)$, i.e., $\frac{\widehat{{\rm var}\left(\mathcal{M}_{n}\right)}}{{\rm var}\left(\mathcal{M}_{n}\right)}\overset{p}{\rightarrow}1$. For expressions of $\widehat{{\rm tr}(\Omega_{n})}$ and $\widehat{{\rm var}\left(\mathcal{M}_{n}\right)}$, refer to Ayyala et al. (2017). Under the null hypothesis, the test statistic is shown to be asymptotically normal. When we define $a_{n}\simeq b_{n}$ if $a_{n}-b_{n}\rightarrow 0$, the power function at significance level $\alpha$ under a local alternative is derived as

[TABLE]

as $n\rightarrow\infty$ where $\Phi$ is the cumulative distribution of standard normal distribution.

The mean in the local alternative satisfies the condition $\mu^{\top}\{\Gamma(a)\Gamma(-a)\}^{1/2}\mu=o\{{\rm tr}(\Omega_{n}^{2})\}$ for all $a\in\mathcal{M}$.

In the proof of asymptotic normality of $\mathcal{T}_{n}$, we used a two-dimensional array argument. The main idea was to divide the array into blocks which were at least $M$ indices apart. The proof was established by showing that (a) these blocks dominate the remainder of terms and (b) the blocks are independent and hence central limit theorem can be used by verifying Lyapunov conditions. However upon further inspection, we see the blocks are not independent, albeit being uncorrelated. Also, we have identified convergence issues for asymptotic power under the rate of increase of $M$ assumed in (3). The main result and numerical studies of Ayyala et al. (2017) remain valid and provide the justification of the use of the test statistic. However, some of the theoretical details of the paper need to be revised.

In this article, we address these issues and provide corrected proofs for the results in Ayyala et al. (2017) as well as the further discussion on the power of the test. The remainder of the paper is organized as follows. In Section 2, a corrected proof for asymptotic normality for the one-sample test statistic is provided. Using the corrected proof, the result for the two-sample case is verified in Section 3. Asymptotic power of the test statistic for the one sample case is presented in Section 4. The notation used in the remainder of the article is made to be consistent with Ayyala et al. (2017). For definitions and additional details on the variables defined, kindly refer to Ayyala et al. (2017).

2 Proof of theorem 3.1 of Ayyala et al. (2017)

In the proof of theorem 3.1 of Ayyala et al. (2017), the authors claim that the variable $B_{ij}$ defined for all $i,j\in\{1,\ldots,k_{n}\}$ (formal definition of $B_{ij}$ is given in Section 2.2) are mutually independent since the process is Gaussian with $M$-dependent stationarity. Using the independence of $B_{ij}$’s, the authors established that the Lyapunov condition, a sufficient condition for the central limit theorem to hold for the leading term of the proposed statistic, and thus the testing statistic was shown to be asymptotically normal. Upon further inspection, we find that the variable $B_{ij}$’s are not mutually independent and hence a dependent central limit theorem is needed to show the asymptotic normality of the proposed statistic.

Here, we present a new proof for the asymptotic normality of the leading term $\sum_{i,j=1}^{k_{n}}B_{ij}$. Following the proof of the theorem 1 of Chen and Qin (2010), we exploit the martingale CLT (Hall and Heyde, 1980, Corollary 3.1). Before the proof, we rewrite some assumptions and notations of Ayyala et al. (2017).

2.1 Assumptions and notations

The authors assume the following:

1. (A1)

$X_{1},\ldots,X_{n}\in\mathbb{R}^{p}$ follow an $M$-dependent strictly stationary Gaussian process with mean $\mu$ and autocovariance structure given by $\Gamma(0),\ldots,\Gamma(M)$. That is, for $h\in\mathcal{M}=\{-M,\ldots,M\}$

[TABLE] 2. (A2)

$p=O(n)$ and $p\rightarrow\infty$. 3. (A3)

$M=O(n^{1/8})$. 4. (A4)

$\Omega_{n}=n{\rm Cov}(\bar{X}_{n})=\sum_{h\in\mathcal{M}}\left(1-\frac{|h|}{n}\right)\Gamma(h)\rightarrow F(0)$ where $F(0)$ is the spectral matrix evaluated at the zero frequency. 5. (A5)

${\rm tr}\left(\Gamma(a)\Gamma(b)\Gamma(c)\Gamma(d)\right)=o\left((M+1)^{-4}{\rm tr}^{2}\left(\Omega_{n}^{2}\right)\right)$ where the rate of decay is uniform for all $a,b,c,d\in\mathcal{M}$.

The naive sample estimator of the autocovariance matrix $\Gamma(h)$ at lag $h\in\mathcal{M}_{+}=\{0,\ldots,M\}$ is denoted by

[TABLE]

with $\widehat{\Gamma}(-h)=\widehat{\Gamma}(h)^{\top}$. To construct an unbiased estimator of ${\rm tr}\left(\Omega_{n}\right)$, let

[TABLE]

Then, the expected value of $\widehat{\gamma}_{n}$ is ${\rm E}\left(\widehat{\gamma}_{n}\right)=\Theta_{n}\gamma$ where $\Theta_{n}\in\mathbb{R}^{\mathcal{M}_{+}\times\mathcal{M}_{+}}$ is a coefficient matrix. Since ${\rm tr}\left(\Omega_{n}\right)=b_{n}^{\rm T}\gamma$ where $b_{n}\in\mathbb{R}^{\mathcal{M}_{+}}$ with $b_{n}(0)=1$ and $b_{n}(h)=2\left(1-\frac{h}{n}\right)$ for $h=1,\ldots,M$,

[TABLE]

is an unbiased estimator of ${\rm tr}\left(\Omega_{n}\right)$ where $\beta_{n}=\Theta_{n}^{-1}b_{n}\in\mathbb{R}^{\mathcal{M}_{+}}$. Therefore, we have

[TABLE]

According to Ayyala et al. (2017), the limiting expression of the variance of $\mathcal{M}_{n}$ is expressed as

[TABLE]

2.2 Theorem 3.1 of Ayyala et al. (2017)

The main theorem of Ayyala et al. (2017) is stated as follows:

Theorem 2.1**.**

(Ayyala et al., 2017, theorem 3.1)* Suppose that the assumptions (A1)$\sim$(A5) and the null hypothesis $H_{0}$ hold. Then, as $n\rightarrow\infty$,*

[TABLE]

To show the asymptotic normality of $T_{n}$, Ayyala et al. (2017) decompose $T_{n}$ as

[TABLE]

where

[TABLE]

and

[TABLE]

and show that $\Delta_{1}\stackrel{{\scriptstyle d}}{{\rightarrow}}N(0,1)$ and $\Delta_{2}\stackrel{{\scriptstyle p}}{{\rightarrow}}0$. To establish the asymptotic normality of $\Delta_{1}$, Ayyala et al. (2017) define the following quantities: an $n\times n$ matrix $\mathcal{A}$ with $(i,j)$th element

[TABLE]

where $\sum\sum_{i,j}\mathcal{A}_{ij}=\bar{X}_{n}^{\rm T}\bar{X}_{n}-\frac{1}{n}{\rm tr}\left(\Omega_{n}\right)$; constants $\alpha\in(0,1)$ and $C>0$ such that $w_{n}=C\cdot n^{\alpha}$, and represent $n=w_{n}k_{n}+r_{n}$ where $0\leq r_{n}<w_{n}$; for $i,j\in\{1,2,\ldots,k_{n}\}$, they define the random variables

[TABLE]

With these, Ayyala et al. (2017) further decompose $\Delta_{1}$ as

[TABLE]

and show that $\Delta_{11}\stackrel{{\scriptstyle d}}{{\rightarrow}}N(0,1)$ and $\Delta_{12}\stackrel{{\scriptstyle p}}{{\rightarrow}}0$. While convergence of $\Delta_{12}$ can be established esasily, asymptotic normality of $\Delta_{11}$ is not straightforward.

2.3 Revised proof

In Ayyala et al. (2017), the authors assumed that the blocks $B_{ij}$’s are independently distributed with unequal covariances. However, they are not independent even though the blocks have zero covariance, ${\rm cov}(B_{ij},B_{i^{\prime}j^{\prime}})=0$ if $(i,j)\neq(i^{\prime},j^{\prime})$. The dependence structure of the blocks should be taken into account in establishing limiting distribution. We provide a new proof for asymptotic normality of $\Delta_{11}$ which takes into account the dependence between the $B_{ij}$ blocks.

Proposition 2.2**.**

Under the assumption of Theorem 2.1, it holds that

[TABLE]

as $n\rightarrow\infty$.

Proof.

For $i=1,\ldots,k_{n}$, let

[TABLE]

Then, $Y_{1},\ldots,Y_{k_{n}}$ are independent and identically distributed Gaussian variables with mean vector 0 and covariance matrix ${\rm Cov}(Y_{i})=\frac{1}{w_{n}-M}\Omega_{w_{n}}$ where $\Omega_{w_{n}}=\sum_{h\in\mathcal{M}}\left(1-\frac{|h|}{w_{n}-M}\right)\Gamma(h)$. We note that for $i\neq j$,

[TABLE]

Following the proof of Theorem 1 of Chen and Qin (2010), define

[TABLE]

and let $\mathcal{F}_{nm}=\sigma(Y_{1},\ldots,Y_{m})$ for $m=2,\ldots,k_{n}$ be the $\sigma$-algebra generated by $Y_{1},\ldots,Y_{m}$. Then we have

[TABLE]

To show the asymptotic normality of $\sum_{i\neq j}B_{ij}$, we need some lemmas.

Lemma 2.3** (square integrable martingale with zero mean).**

For each $n$, $(S_{nm},\mathcal{F}_{nm})_{m=2}^{k_{n}}$ is the sequence of zero mean and a square integrable martingale.

Proof of lemma 2.3.

Since $Y_{i}$’s are i.i.d. Gaussian random variables with zero mean, $S_{nm}$ has zero mean and is square integrable for any $n,m$. We observe

[TABLE]

Therefore, $(S_{nm},\mathcal{F}_{nm})_{m=2}^{k_{n}}$ is a zero mean and square integrable martingale sequence. ∎

Lemma 2.4** (analogous condition on the conditional variance).**

Let $\sigma_{n}^{2}=\frac{2k_{n}(k_{n}-1)(w_{n}-M)^{2}}{n^{4}}{\rm tr}\left(\Omega_{w_{n}}^{2}\right)$. Then,

[TABLE]

Proof of lemma 2.4.

We note that

[TABLE]

Define $\eta_{n}=\sum_{j=2}^{k_{n}}{\rm E}\left[V_{nj}^{2}|\mathcal{F}_{n(j-1)}\right]$, then its first moment is

[TABLE]

The second moment of $\eta_{n}$ is

[TABLE]

and thus the variance of $\eta_{n}$ is

[TABLE]

By assumption (A5), it holds that ${\rm tr}\left(\Omega_{w_{n}}^{4}\right)=o\left({\rm tr}^{2}\left(\Omega_{w_{n}}^{2}\right)\right)$. Thus, we have

[TABLE]

This completes the proof. ∎

Lemma 2.5** (conditional Lindeberg condition).**

For any $\epsilon>0$,

[TABLE]

Proof of lemma 2.5.

We note that

[TABLE]

It suffices to show

[TABLE]

We have that

[TABLE]

This completes the proof. ∎

We have thus established the sufficient conditions for martingale Central Limit Theorem. By the Corollary 3.1 of Hall and Heyde (1980), Lemmas 2.3, 2.4 and 2.5,

[TABLE]

We define

[TABLE]

if $a_{n}=b_{n}(1+o(1))$ for $a_{n}>0$ and $b_{n}>0$. Note that $\frac{\sigma_{n}}{\sqrt{Var(M_{n})}}\rightarrow 1$ since $\frac{\sigma_{n}^{2}}{Var(M_{n})}-1\asymp\frac{tr(\Omega_{w_{n}}^{2})-tr(\Omega_{n}^{2})}{tr(\Omega_{n}^{2})}$ and

[TABLE]

where $\leq$ in (16) is due to $\Gamma(i)\Gamma(j)=o(\sqrt{n}tr(\Omega_{n}^{2}))$ from Lemma C.1 in the appendix of Ayyala et al. (2017) and the equality in (17) is obtained by taking $w_{n}\geq(M+1)\sqrt{n}$. Also, by the classical central limit theorem, it holds that $\frac{\sum_{i=1}^{k_{n}}B_{ii}}{\sqrt{k_{n}{\rm Var}(B_{11})}}\stackrel{{\scriptstyle d}}{{\rightarrow}}N(0,1)$. Since

[TABLE]

we have

[TABLE]

Therefore, we get the desired result

[TABLE]

∎

3 Two Sample Test

By construction of the blocks, we have $\frac{w_{n_{g}}-M}{n_{g}}\asymp\frac{1}{k_{n_{g}}}$ from ${n_{g}}=k_{n_{g}}w_{n_{g}}+r_{n_{g}}$ with $w_{n_{g}}$ satisfying $M=o(w_{n_{g}})$ for $g=1,2$ where $\asymp$ is defined in (15). As done in one sample problem, if we define

[TABLE]

where $Y_{gi}=\frac{1}{w_{n_{g}}-M}\sum_{k=(i-1)w_{n_{g}}+1}^{iw_{n_{g}}-M}X_{gk}$, then two sample test statistic is

[TABLE]

where

[TABLE]

and $\Delta_{12}^{(l)}$ for $l=1,2$ is defined as in one sample case.

We change the condition (14) in Ayyala et al. (2017) to the following:

[TABLE]

As in one sample case, we can show

[TABLE]

We now prove the asymptotic normality of $T_{n}$. Since $Y_{11},Y_{12},\ldots,Y_{1k_{n_{1}}}$ and $Y_{21},Y_{22},\ldots,Y_{2k_{n_{2}}}$ are $k_{n_{1}}+k_{n_{2}}$ observational vectors with $E(Y_{gi})=\mu_{g}$ and $Cov(Y_{gi})=\frac{1}{(w_{n_{g}}-M)}\Omega^{(g)}_{w_{n_{g}}}$ for $g=1,2$, we can apply the martingale C.L.T. in Chen and Qin (2010) to independent $k_{n_{1}}+k_{n_{2}}$ observations. Therefore we obtain

[TABLE]

where

[TABLE]

since $M=o(w_{n_{g}})$ and $k_{n_{g}}w_{n_{g}}\asymp n_{g}$. Furthermore, we can also show the asymptotic equivalence of ${\rm var}(T_{n})$ and ${\rm var}\left(\mathcal{M}_{n}\right)$ which is stated in the following lemma.

Lemma 3.1**.**

Under the conditions given in Ayyala et al. (2017), we have ${\rm var}(T_{n})\asymp{\rm var}\left(\mathcal{M}_{n}\right)$.

Proof.

By using the following facts:

• •

when $n_{1}$ and $n_{2}$ have the same order except constant factor, ${\rm var}\left(\mathcal{M}_{n}\right)\asymp tr(\frac{\Omega_{n_{1}}}{n_{1}}+\frac{\Omega_{n_{2}}}{n_{2}})^{2}$ has the same order as $\frac{1}{n^{2}}tr(({\Omega_{n_{1}}}+{\Omega_{n_{2}}})^{2})$ except some constant factor. This can be show using $tr(\Omega_{n_{l}}^{2})>0$ for $l=1,2$ and $tr(\Omega_{n_{1}}\Omega_{n_{2}})\geq 0$

• •

when $p=O(n)$, $\Gamma(i)\Gamma(j)=\sqrt{p}o(tr(\Omega_{n_{1}}+\Omega_{n_{2}})^{2})$ in Lemma C.1 in Appendix in Ayyala et al. (2017),

we can derive

[TABLE]

where the last equality holds if $w_{n}=[n^{\alpha}]$ for $\frac{7}{8}\leq\alpha<1$. ∎

Combining (18), (21) and Lemma 3.1, we have

[TABLE]

Additionally, since we have $\frac{\widehat{{\rm var}\left(\mathcal{M}_{n}\right)}}{{\rm var}\left(\mathcal{M}_{n}\right)}\stackrel{{\scriptstyle p}}{{\rightarrow}}1$ from Ayyala et al. (2017), we finally have the asymptotic normality of the proposed test for two sample case in Ayyala et al. (2017):

[TABLE]

4 Asymptotic power

To calculate the asymptotic power of $\mathcal{T}_{n}$ under the local alternative condition in equation (11), define $X_{t}=\mu+Y_{t},t=1,\ldots,n$ where $Y_{t}$ has mean zero and autocovariance structure given by $\{\Gamma(h)\}$. Define the test statistic calculated using $\mathbf{X}=\{X_{1},\ldots,X_{n}\}$ and $\mathbf{Y}=\{Y_{1},\ldots,Y_{n}\}$ as $\mathcal{T}_{n}(\mathbf{X})$ and $\mathcal{T}_{n}(\mathbf{Y})$ respectively. Similarly, we also define $\mathcal{M}_{n}(\mathbf{X})$ and $\mathcal{M}_{n}(\mathbf{Y})$ corresponding to $\mathcal{M}_{n}$ defined in Section 3. By Theorem 3.1, we have asymptotic normality of $\mathcal{T}_{n}(\mathbf{Y})$,

[TABLE]

For any non-zero $\mu$ satisfying the local alternative condition, we have

[TABLE]

By Theorem 3.2, the variance estimator is ratio-consistent even under the local alternative, hence $\widehat{{\rm var}(\mathcal{M}_{n}(\mathbf{X}))}/{\rm var}(\mathcal{M}_{n}({\bf Y}))\stackrel{{\scriptstyle p}}{{\rightarrow}}1$. The matrix $\Omega_{n}$ is the second central moment of the data and is hence independent of $\mu$. Thus, we have $\Omega_{n}(\mathbf{X})=\Omega_{n}(\mathbf{Y})=\Omega_{n}$. Expressing $\mathcal{M}_{n}(\mathbf{X})$ in terms of $\mathbf{Y}$, we have

[TABLE]

The second term, $\mu^{\top}\overline{Y}_{n}$ is normally distributed with $E(\mu^{\top}Y)=\mu^{\top}0=0$ and variance ${\rm var}({\mu^{\top}\overline{Y}_{n}})=\frac{1}{n}\mu^{\top}\Omega_{n}\mu$. This gives

[TABLE]

where the last equality follows from Proposition 3.1. Writing $\Omega_{n}$ in terms of the autocovariance matrices $\Gamma(h),h\in\mathcal{M}$, we have the ratio as a linear combination of $\mu^{\top}\Gamma(h)\mu,h=0,\pm 1,\ldots,\pm M$.

For each $h$, we consider the singular value decomposition (SVD) of $\Gamma(h)$, $\Gamma(h)=U\Lambda V^{\top}$, where $\Lambda$ is the diagonal matrix with positive square roots of eigenvalues of $\left[\Gamma(h)\Gamma(h)^{\top}\right]$. It yields $\Gamma(h)\Gamma(-h)=U\Lambda^{2}U^{\top}$ which gives $U\Lambda U^{\top}=\left(\Gamma(h)\Gamma(-h)\right)^{1/2}$. Thus,

[TABLE]

where the inequality comes from the Cauchy-Schwarz inequality and

[TABLE]

The upper bound on $\mu^{\top}\Omega_{n}\mu$ is given by

[TABLE]

In sequel, the ratio in (26) is bounded by

[TABLE]

Ayyala et al. (2017) make the assumption (see Eqn. (11) of Ayyala et al. (2017))

[TABLE]

and this makes the summands in (29) are all in the order of $o(1)$ and the summation (29) converges to zero if $M$ is finite. In this article, we assume $M=O(n^{1/8})$ and change the condition (30) accordingly to

[TABLE]

In (31), if $M$ is bounded, the condition (31) equals to the condition (30) that is in Ayyala et al. (2017). When $M$ has the order of $n^{1/8}$, the order in (31) becomes $o\left(n^{-9/8}{\rm tr}\left(\Omega_{n}^{2}\right)\right)$. With the assumptions $M=O(n^{1/8})$ and (31), we have the variance of $\mu^{\top}\overline{Y}_{n}/\sqrt{{\rm var}{\mathcal{M}_{n}({\bf Y})}}$ converging to zero. Further, this quantity has expected value zero, we claim that it converges to zero in probability.

Combining asymptotic normality of $\mathcal{M}_{n}(\mathbf{Y})$ and the result above, we can show that

[TABLE]

5 Concluding remarks

We conclude the note with a few additional remarks. First, for practical use, the denominator of the test statistic in of Theorem 8 is replaced with the ratio consistent variance estimator given in Theorem 3.2 of Ayyala et al. (2017). For asymptotic normality using ratio consistent estimator, uncorrelatedness of $B_{ij}$s is sufficient and hence two main statements (Theorem 3.1 and 3.2 of their paper) of Ayyala et al. (2017) remain valid and also the numerical study presented in Ayyala et al. (2017) is meaningful. Second, for asymptotic power calculation, Ayyala et al. (2017) assert that asymptotic normality of the statistic $\frac{M_{n}-\mu^{\rm T}\mu}{\sqrt{\widehat{{\rm Var}\left(M_{n}\right)}}}$ follows from ${\rm E}\left(M_{n}\right)=\mu^{\rm T}\mu$. However, in order to have this result, the local alternative condition needs to be modified. The power function holds only if the mean belongs to the local alternative mentioned in (31), instead of the condition in equation (11) in Ayyala et al. (2017). To show that the statistic is asymptotically normal under the alternative hypothesis, more complicate proof is required. Finally, we take this opportunity to list some other typos in Ayyala et al. (2017):

On pp. 138, the identity $\mathrm{E}\left[n\bar{X}_{n}^{\top}\bar{X}_{n}-{\rm tr}(S_{n})\right]=\mu^{\top}\mu$ should be ${\rm E}\left[\bar{X}_{n}^{\top}\bar{X}_{n}-n^{-1}{\rm tr}(S_{n})\right]=\mu^{\top}\mu$. 2. 2.

On pp. 139, the expected value of Euclidean norm of sample mean should be

[TABLE] 3. 3.

The value of $b_{n}(0)$ is mentioned on pp. 139 do not have a factor of $1/n$ multiplied. It should be changed by mutiplying by $n$, i.e. $b_{n}(0)=1$ and $b_{n}(i)=2\{1-(i-1)/M\}$ for $i\in\{2,\ldots,M+1\}$. The expected value of $M_{n}=\bar{X}_{n}^{\top}\bar{X}_{n}-\frac{1}{n}\widehat{{\rm tr}(\Omega_{n})}$ is $\mu^{\top}\mu$. 4. 4.

The value of $\pi_{n}$ presented on pp. 139 should be corrected as

[TABLE]